The outer edge of the grooved area of a long-playing record is at a radial dista

Question

The outer edge of the grooved area of a long-playing record is at a radial distance 30 cm from the center; the inner edge is at a radial distance of 18 cm. The record rotates at 6.2 rev/min. The needle of the pick-up arm takes 1.7 minutes to move uniformly from the outer edge to the inner edge.

What is the radial speed of the needle? m/s

What is the speed of the outer edge relative to the needle? m/s

What is the speed of the inner edge relative to the needle? m/s

Suppose the phonograph is turned off, and the record uniformly and stops rotating after 10 s. What is the angular acceleration? rad/s2

Explanation / Answer

1) Since it takes 1.7 minutes for the needle to move from outer edge to inner edge, and the radial distance from the outer edge to the inner edge is 30cm - 18cm = 12 cm, the radial speed of the needle is 12cm/1.7min = 1.18*10^-3 m/s. 2) The angular velocity is omega = 6.2 rev/min = 0.65 rad/s. So the speed of the outer edge is v_outer = omega * R_outer = 0.65 rad/s * 30*10^-2 m = 0.19 m/s. 3) The speed of the inner edge is v_inner = omega * R_inner = 0.65 rad/s * 18*10^-2 m = 0.12 m/s. 4) The angular acceleration is alpha = (omega_final - omega_init) / t = (0-0.65rad/s) / 10s = -0.065 rad/s^2.

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